Question

spiral review 4 - grade in sis which of the following has a value between 10 and 11? $\sqrt{143}$ $\sqrt{90}$ $\sqrt{102}$ $\sqrt{167}$

Explanation:

Step1: Find squares of 10 and 11

We know that \(10^2 = 100\) and \(11^2 = 121\). So we need to find a number whose square root is between 10 and 11, which means the number inside the square root (radicand) should be between \(10^2 = 100\) and \(11^2 = 121\).

Step2: Check each option

• For \(\sqrt{143}\): Since \(143>121\) (because \(11^2 = 121\)), \(\sqrt{143}>11\).
• For \(\sqrt{90}\): Since \(90 < 100\) (because \(10^2 = 100\)), \(\sqrt{90}<10\).
• For \(\sqrt{102}\): Since \(100<102<121\), \(\sqrt{100}<\sqrt{102}<\sqrt{121}\), which simplifies to \(10 < \sqrt{102}<11\).
• For \(\sqrt{167}\): Since \(167>121\), \(\sqrt{167}>11\).

Answer:

\(\boldsymbol{\sqrt{102}}\) (the option with \(\sqrt{102}\))